# Let x=4 be a directrix to an ellipse whose centre is at the

Question:

Let $x=4$ be a directrix to an ellipse whose centre is at the

origin and its eccentricity is $\frac{1}{2}$. If $P(1, \beta), \beta>0$ is a point on

this ellipse, then the equation of the normal to it at $P$ is :

1. (1) $4 x-3 y=2$

2. (2) $8 x-2 y=5$

3. (3) $7 x-4 y=1$

4. (4) $4 x-2 y=1$

Correct Option: , 4

Solution:

$\frac{a}{e}=4 \Rightarrow a=4 \times \frac{1}{2}=2$

Now, $b^{2}=a^{2}\left(1-e^{2}\right)$

$\Rightarrow b^{2}=4\left(1-\frac{1}{4}\right)=4 \times \frac{3}{4}=3$

So, equation $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$

$\Rightarrow 3 x^{2}+4 y^{2}=12$ .....(i)

Now, $P(1, \beta)$ lies on it

$\Rightarrow 3+4 \beta^{2}=12 \Rightarrow \beta=\frac{3}{2}$

So, equation of normal at $P\left(1, \frac{3}{2}\right)$

$\Rightarrow \frac{a^{2} x}{1}-\frac{b^{2} y}{3 / 2}=a^{2}-b^{2} \Rightarrow 4 x-2 y=1$